Let's assume that there is process $Z_t$: $$Z_t = \exp-(\sigma(T-t)W_t+\sigma \int_0^t W_sds + \int_0^Tf(0,u)du + \int_0 ^t\int_s^T\alpha(s,u)duds) $$ where $f(t,T)$ means instantaneous forward rate, and $\alpha$ is a drift, bounded deterministic function. (This is taken from Baxter and Rennie Financial Calculus). I want to find SDE for above process. I know that I need to use Ito's lemma, assuming that $Z_t = e^{-Y_t}$, where $Y_t$ is expression in bracket. However, I am not sure how to find $dY_t$, as I am having problem with $\sigma \int_0^t W_sds$ expression (I think $\int_0^Tf(0,u)du$ $ $ will be a constant, so we can omit it in SDE for $Y_t$). The remaining two expressions for $Y_t$ are pretty easy to deal with.

Edit: Thanks @Did. So $$dY_t=\sigma(T-t)dW_t + \left( \sigma W_t + \int_t^T\alpha(s,u)du\right) dt$$ Now, if $F(t,x)=e^{-x}$, then $F_t=0$, $F_x=-F(t,x)$ and $F_{xx}=F(t,x)$. Using Ito's formula, we have $$dZ_t=Z_t\left(-\sigma(T-t)dW_t \, + \left(\frac{1}{2}\sigma^2(T-t)^2-\sigma W_t - \int_t^T\alpha(s,u)du\right)dt \right)$$ But in the book the SDE does not contain $-\sigma W_t$. Am I missing something here? $f$ function is defined $f(t,T)= \sigma W_t + f(0,T) + \int_0^t \alpha (s,T)ds$, so $\int_0^Tf(0,u)du$ should be some constant and should not affect $dY_t$.

Edit2: As pointed out by @RhysSteele, I missed $-\sigma W_t dt$ expression when calculating $d(\sigma (T-t) W_t)$. Correct expression for $dY_t$ is $$dY_t=\sigma(T-t)dW_t + \left(\int_t^T\alpha(s,u)du\right) dt$$ and $dZ_t$ is then equal $$dZ_t=Z_t\left(-\sigma(T-t)dW_t \, + \left(\frac{1}{2}\sigma^2(T-t)^2 - \int_t^T\alpha(s,u)du\right)dt \right)$$

  • $\begingroup$ $$d\left(\sigma \int_0^t W_sds\right)=\sigma W_tdt$$ $\endgroup$
    – Did
    Aug 15, 2018 at 19:06
  • $\begingroup$ Your expression for $dY_t$ is incorrect. Im guessing that what went wrong is that you didn't find the correct form for $d \bigg(\sigma(T-t)W_t \bigg) = \sigma (T-t)dW_t - \sigma W_t dt$ (which follows from Ito's formula). $\endgroup$ Aug 15, 2018 at 20:35
  • $\begingroup$ Of course you are right @RhysSteele, thanks for pointing this out $\endgroup$
    – siwy9
    Aug 15, 2018 at 20:48

1 Answer 1


Using hints in comments I will post full solution here.

We have $$dY_t=\sigma(T-t)dW_t + \left( - \sigma W_t + \sigma W_t + \int_t^T\alpha(s,u)du\right) dt =\sigma(T-t)dW_t + \left(\int_t^T\alpha(s,u)du\right) dt$$ Then, for Ito lemma, we use function $F(t,x)=e^{-x}$.

The partial derivatives are: $F_t=0$, $F_x=-F(t,x)$ and $F_{xx}=F(t,x)$.

So $dZ_t$ has following formula: $$dZ_t=Z_t\left(-\sigma(T-t)dW_t \, + \left(\frac{1}{2}\sigma^2(T-t)^2 - \int_t^T\alpha(s,u)du\right)dt \right)$$


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